Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)

Answer 1

Construction: Draw BQ ⊥ AC and DR ⊥ AC

Proof:
L.H.S
 = ar (Δ APB ) × ar (ΔCP) 

 = `1/2[ ( AP xx BQ )] xx (1/2 xx PC xx DR)`

 =`( 1/2 xx PC xx BQ )xx (1/2 xx AP xx DR)` 

 = ar (Δ BPC )  × ar (APR)

 = RHS

 ∴ LHS = RHS

Hence proved.